# Moniod with a unit

• Oct 21st 2009, 05:45 PM
sfspitfire23
Moniod with a unit
Hello!

If M is a finite monoid and $\displaystyle au=bu$ in M implies that $\displaystyle a=b$, show that $\displaystyle u$ is a unit.

with hints: If $\displaystyle M=\{a_1,\cdots a_n\}$ show that $\displaystyle a_1u,\cdots , a_nu$ are distinct.

Attempt:

If $\displaystyle M=\{a_1,\cdots a_n\}$ then by definition it has an identity element $\displaystyle e$. Then, because M is finite, there will be an element which will map to the identity of M. Each $\displaystyle a_1u\cdots a_nu$ will be distinct as for every $\displaystyle a_n$, u will map it back to itself. Thus, $\displaystyle au=bu$ which, because $\displaystyle u$ will map $\displaystyle a$ to itself $\displaystyle a(e)=b(e)$ and $\displaystyle a=b(e)$ so $\displaystyle u$ must be a unit.

Thanks guys!
• Oct 22nd 2009, 03:50 AM
clic-clac
Hi

What you have to prove is: $\displaystyle \exists b\in M\ \text{s.t.}\ bu=e.$

To show the hint, just use that $\displaystyle au=bu\Rightarrow a=b,$ (for instance prove it by contraposition).

Assume you've proved it. Show that $\displaystyle M=\{a_1u,...,a_nu\}$ (using a property of finite sets)

Therefore $\displaystyle e\in M$ means $\displaystyle e\in \{a_1u,...,a_nu\}$. Can you conclude?
• Oct 22nd 2009, 09:10 PM
sfspitfire23
would it suffice to create a cayley table with elements 1,a,b and show that a and b both have an inverse? ugh, where would i go from there?

thanks, really appreciate it
• Oct 22nd 2009, 11:12 PM
clic-clac
I don't really understand, why just $\displaystyle e,a,b$ in a Cayley table. $\displaystyle ab$ can be different from $\displaystyle e,a$ or $\displaystyle b.$

What the proof says is : the condition you have implies that the map $\displaystyle M\rightarrow M: a\mapsto au$ is injective and then surjective (because it's an injective map between sets of same finite cardinality) therefore $\displaystyle e$ has an antecedant, which is some $\displaystyle b\in M$ such that $\displaystyle bu=e,$ and we're done.